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A 4460 lb Ford Explorer traveling 35 mph has a head on collision with a 2750 lb Toyota Corolla, also traveling 35 mph.

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{excerpt:hidden=true}Compare the forces on the occupants of two cars in a 1-D totally inelastic collision.{excerpt}
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{card:label=Part A}
h2. Part A

Assuming that the automobiles become locked together during the collision, what is the speed of the combined mass immediately after the collision?

h4. Solutions

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}Explorer plus Corolla as [point particles|point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}[Impulse|impulse] resulting from [external influences|external force] will be neglected, as we assume that the collision is instantaneous.{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Momentum and External Force].{cloak}

{toggle-cloak:id=appa} *Approach:*  

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{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diaga}
We begin by sketching the situation and defining a [coordinate system|coordinate system].

|!CollisionInit.jpg|height=175!|!CollisionFin.jpg|height=175!|
||Initial State||Final State||

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{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

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Since we assume that [external|external force] [forces|force] are negligible during the collision, we set the [external|external force] [impulse|impulse] to zero which gives:

{latex}\begin{large}\[ p^{TC}_{x,i} + p^{FE}_{x,i} = p^{system}_{x,f} \]\end{large}{latex}

or, in terms of the masses:

{latex}\begin{large}\[ m^{TC}v^{TC}_{x,i} + m^{FE}v^{FE}_{x,i} = (m^{TC}+m^{FE})v_{x,f} \]\end{large}{latex}

which gives:

{latex}\begin{large}\[ v_{x,f} = \frac{m^{TC}v^{TC}_{x,i} + m^{FE}v^{FE}_{x,i}}{m^{TC}+m^{FE}} = \mbox{3.71 m/s} = \mbox{8.3 mph}\]\end{large}{latex}

{warning}Remember that in our [coordinate system|coordinate system], the Corolla has a negative x-velocity before the collision.{warning}

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Part B

Find the impulse that acted on each of the vehicles during the collision.

Solution

Part C

Assuming the collision lasted for 0.060 seconds, find the time-averaged force exerted on each vehicle.

Solution

Part D

Suppose a 75 kg person in each vehicle underwent the same change in velocity as their automobile in the same amount of time. Find the time-averaged force exerted on these people.

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{card:label=Part B}

h2. Part B

Find the [impulse|impulse] that acted on each of the vehicles during the collision.

h4. Solution

{toggle-cloak:id=sysb} *Systems:* {cloak:id=sysb} Corolla and Explorer as *separate* [point particle] [systems|system].{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}The [impulse|impulse] on each vehicle from the other is assumed to be the dominant [interaction|interaction] during the collision.  Because we are now considering the vehicles separately, these are now [external|external force] [impulses|impulse].{cloak}

{toggle-cloak:id=modb} *Model:* {cloak:id=modb}[Momentum and External Force].{cloak}

{toggle-cloak:id=appb} *Approach:* 

{cloak:id=appb}

With the results of Part A it is straightforward to calculate the [impulse|impulse] on the Explorer due to the collision.  To be specific, we label the impulse "EC" to remind ourselves the [impulse|impulse] on the Explorer is provided by the Corolla.

{latex}\begin{large}\[ J^{EC}_{x} = p^{E}_{x,f}-p^{E}_{x,i} = m^{E}(v^{E}_{x,f}-v^{E}_{x,i}) = (\mbox{2030 kg})(\mbox{3.71 m/s}-\mbox{15.6 m/s}) = -\mbox{24000 kg m/s} \] \end{large}{latex}

Similarly, for the Corolla:

{latex}\begin{large}\[ J^{CE}_{x} = p^{C}_{x,f}-p^{C}_{x,i} = m^{C}(v^{C}_{x,f}-v^{C}_{x,i}) = (\mbox{1250 kg})(\mbox{3.71 m/s}+\mbox{15.6 m/s}) = +\mbox{24000 kg m/s} \] \end{large}{latex}

{note}Again, it is important to note that the Corolla's initial x-velocity is negative in our chosen [coordinate system|coordinate system].{note}

{tip}It is no coincidence that _J{_}{^}EC^ = -- _J{_}{^}CE^.  The relationship is guaranteed by [Newton's 3rd Law|Newton's Third Law].{tip}

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{card}
{card:label=Part C}

h2. Part C

Assuming the collision lasted for 0.060 seconds, find the [time-averaged|impulse] [force|force] exerted on each vehicle.

h4. Solution

{toggle-cloak:id=sysc} *Systems:* {cloak:id=sysc} Corolla and Explorer as *separate* [point particle] [systems|system].  {cloak}

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc} The [external|external force] [force|force] on each vehicle from the other is assumed to be the dominant [interaction|interaction] during the collision.{cloak}

{toggle-cloak:id=modc} *Model:* {cloak:id=modc}[Momentum and External Force].{cloak}

{toggle-cloak:id=appc} *Approach:*  

{cloak:id=appc}
From the definition of [impulse|impulse] we know:

{latex}\begin{large}\[ J^{EC}_{x} = \int F^{EC}_{x}\:dt \equiv \bar{F}^{EC}_{x} \Delta t\]\end{large}{latex}

so the time-averaged [force|force] exerted on the Explorer is:

{latex}\begin{large}\[\bar{F}^{EC}_{x} = \frac{J^{CE}_{x}}{\Delta t} = -\mbox{400,000 N} \]\end{large}{latex}

Similarly, the time-averaged [force|force] exerted on the Corolla is:

{latex}\begin{large}\[\bar{F}^{CE}_{x} = \frac{J^{EC}_{x}}{\Delta t} = \mbox{400,000 N}\]\end{large}{latex}

{tip}Again, the relationship between the two [forces|force] is guaranteed by [Newton's 3rd Law|Newton's Third Law].{tip}

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{card:label=Part D}


h2. Part D

Suppose a 75 kg person in each vehicle underwent the same change in [velocity|velocity] as their automobile in the same amount of time.  Find the [time-averaged|impulse] [force|force] exerted on these people.

{toggle-cloak:id=sysd} *Systems:* {cloak:id=sysd}First, the Corolla and Explorer as separate [point particle] systems, then the passengers as *separate* [point particle] systems.{cloak}

{toggle-cloak:id=intd} *Interactions:*  {cloak:id=intd} Each vehicle is subject to a collision [force|force] from the other.  The passengers are each subject to some decelerating [force|force], possibly a seatbelt or airbag.{cloak}

{toggle-cloak:id=modd} *Model:* {cloak:id=modd}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appd} *Approach:*  

{cloak:id=appd}
The passengers clearly are not subject to the same [force|force] as their vehicles.  Rather, they are subject to the same [accelerations|acceleration].  (They are -- hopefully -- strapped into their automobiles, so that whatever happens to their vehicle's motion happens to their motion as well.)  Thus, our first goal is to determine the vehicles' [accelerations|acceleration].  This is easily accomplished using the results of Part C in [Newton's 2nd Law|Newton's Second Law].

{latex}\begin{large}\[ a^{E}_{x}= \frac{F^{EC}_{x}}{m^{E}} = -\mbox{200 m/s}^{2}\]
\[ a^{C}_{x} = \frac{F^{CE}_{x}}{m^{C}} = \mbox{320 m/s}^{2} \]\end{large}{latex}

{tip}The Corolla's [acceleration|acceleration] should clearly be larger, even though the beginning and ending speeds of the Explorer and Corolla are the same.  The reason is that the Corolla has _changed direction_.  Thus, it effectively dropped all the way to zero from 35 mph and then _accelerated_ back up to 8.3 mph the other way.  The Explorer simply dropped its forward speed from 35 mph to 8.3 mph.{tip}

{info}Note (discounting safety measures like airbags) the SUV driver's acceleration would be about 20 [g's|gee], near the limit of human endurance.  The Corolla's driver would experience almost 33 [g's|gee].{info}

With these [accelerations|acceleration], we can find the [force|force] on each driver:

{latex}\begin{large}\[ F^{E,driver}_{x} = m^{driver}a^{E}_{x} = -\mbox{15000 N} = -\mbox{3,300 lbs}\]
\[ F^{C,driver}_{x} = m^{driver}a^{C}_{x} = \mbox{24000 N} = \mbox{5,400 lbs}\]\end{large}{latex}

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